Regarding quantum physics, please help me look at the Chinese translation 2. Thank you!

Isn't that part of Richard Dawkins' speech? Scientific squirrels will have ready-made translations.

Science often seriously challenges common sense.

What is it that makes us think of anything? And does that explain what we can expect? Is there something in the universe that is forever beyond our comprehension, but at the same time not beyond the comprehension of some higher intelligent beings?

Is there anything in the universe that in principle transcends all minds, no matter how advanced they may be?

The history of science is a long string of intense brainstorming as new generations of theories essentially embrace more bizarre aspects of the universe. We are now so accustomed to the idea that the earth is rotating – not the sun moving in the sky – that it's hard to imagine what a shocking intellectual revolution it once was. After all, everything is so obvious, the Earth is big and stationary, and the Sun is small and moving.

Tell me", he asked a friend, "why is it natural for people to always say that the sun revolves around the earth and not on its own rotation?" His friend said, "Well, obviously, because it looks like the sun is going around the earth."

To this, he said: "Okay, so to make the Earth look like it's rotating, what should it look like? ”

I used unicell's translation because I thought it was better than Squirrel would translate. In addition, I note that you have made some changes in the original text, so I have made corresponding adjustments in the translation. In the end, your reference to Wittgenstein's story is omitted from his name, so it's better to change "he asked a friend" to "a man asked a friend" so that it is less obtrusive.

For more information on this question, you can see "The History of Quantum Mechanics" (a lot of literature), which is the best (and most understandable) popular book in my opinion.

Let me explain briefly:

The state of a quantum mechanical system is largely influenced by the observer. This is mainly done by measuring behavior. Therefore, when the measurement behavior is not done, the system is reversible and seems to be called a u-process.

If the measurement occurs, the system immediately undergoes irreversible changes, which seems to be called the r process.

I also know a little bit about it, I hope it will help you.

The result of the measurement is a quantum transition in the system, such as double-slit diffraction of electrons or other microscopic particles, and the electrons hit the position of the stripes on the screen with a certain probability, so it is impossible to determine which stripes will be hit in this process. The screen is a measurement of the position of electrons, once the electrons hit the screen, they will leave a trace, and will no longer appear in space with a certain probability as before, that is, a quantum jump has occurred, and the jump to a certain state.

The term reversible, irreversible is not very good.

(1) This is the matrix representation of the sx, sy, and sz components with spin of 1. Knowing sz, the above results can be obtained by simple derivation of the ascending and lowering operator s.

2) The initial state is |11>=(1,0,0) (transposed), the Hamiltonian of the system is in the x-direction, and you need to find the eigenstates under sx respectively.

sx=1>=(1 2, root number 2, 1 2), sx=0>=....sx=-1>=...Do the math yourself) and then project it over|11>=a1|sx=1>+a2|sx=0>+a3|sx=-1>, an=<11|sx=n> this should count.

So the wave function of time-dependent evolution is |t>=a1|sx=1>e^(-iet)+a2|sx=0>+a3|sx=-1>e^(iet), e=gbh/2π

3)|1-1>=(0,0,1)(transpose), found in |The probability of a 1-1 > is ||^2。

For the sake of writing convenience, order bai

du(0)= (x,0), t)= (x,0), zhin= n(x), dao by default represents the sum of the genus from 0 to .

1) Because.

n|m>=δnm, so 1= < 0)|ψ0)>=n^2*σc^2n，n^2=1/σc^2n，n=...

2) h|ψn>=(n+1/2) ωn>=en|n>, so (t)=n c n*e (ient) n

3) The probability is |<ψt)|ψ0)>|2. Do the math yourself, use < n|m>=δnm, which should be a good calculation.

4) h|ψn>=(n+1/2) ωn>=en|n>, so =< (0)|h|ψ(0)>=n^2*σc^2n*en

There is a problem in the middle of the method, the average value of the last px is not 1, and the correct process is as follows:

Finally, in the second step, where p is the differential operator, it is broken down into two terms, one p acting on x and one p acting on the subsequent wave function. However, I still don't know how to calculate it next, and I feel that your proposition should not be true for all systems.

This is the most basic adiabatic perturbation problem.

The original wave function is the eigenwave function of a one-dimensional infinitely deep potential well. Then when the potential well becomes 4 times, the wave function is still the same, note that it only exists in the segment (0 x a), and (a x 3a) the wave function is 0. The original wave function is set to (0), and the wave function is set to the new eigenwave function (i.e., the eigenwave function of the 4 times wide infinitely deep potential well)|φ0)>=cn|n >, so the probability is equal to.

cn|^2=||2, n=0, 1 correspond to the ground state and the first excited state probability, respectively.

The operator p=m ih [x,h], and by this you can calculate the average value of p, and you will use the state it gives you, and you will calculate that it is equal to (en-en) and a coefficient, because en-en is equal to 0, so no matter what the others are, the average value of your p is 0